# What type of equation is this one?

1. Apr 27, 2014

### Beerdrinker

Hi,

I am struggling with a problem, where in the middle of my calculations I need to determine a function $f(\alpha, \beta; x)$, namely the function of $x$ parametrized by $\alpha$ and $\beta$, from the following equation

$$f\left(\alpha, \beta;\frac{1}{x}\right) = x^4f(\gamma, \delta; x).$$

So the $f(1/x)$ on the LHS and $f(x)$ on the RHS may differ in parametrization. Somehow I found that this equation admits the following solution

$$f(\alpha, \beta; x) = \left(\alpha + \beta x^a\right)^b$$

where $a\cdot b = -4$. Although, this is probably not the most general solution.

So my question is what type of equation this one is? And how one should approach such equation?

Last edited: Apr 27, 2014
2. Apr 27, 2014

### disregardthat

Some clarifying questions:

Is x a non-zero real number?

Are $\alpha, \beta$ real numbers?

Are $\gamma$ and $\delta$ real number constants (independent of $\alpha, \beta, x$)?

In your example, you're assuming $\gamma = \alpha = \delta = \beta$?

3. Apr 27, 2014

### Beerdrinker

$f(x)$ is well defined and restricted to either $x>0$ or $x<0$.
$\alpha$, $\beta$, $\gamma$, $\delta$ in general may be complex and independent of $x$ (they parametrize function $f(x)$).
However $\gamma$, $\delta$ may be functions of $\alpha$, $\beta$.

In the example I provided previously we have

$$f\left(\alpha, \beta; x\right) = \left(\alpha + \beta x^{a} \right)^b \Rightarrow f\left(\alpha, \beta; \frac{1}{x}\right) = \left(\alpha + \beta x^{-a} \right)^b = x^{-ab}\left(\alpha x^a + \beta\right)^b = x^{4}f(\beta, \alpha; x)$$
Where the assumption $ab = -4$ was used. So in this case $\alpha = \delta$ and $\beta = \gamma$ and are not neccesarily equal.

Sorry for not making this clear enough earlier.

Last edited: Apr 27, 2014
4. May 4, 2014

Anyone?

5. May 4, 2014